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joja [24]
4 years ago
7

You are attending the State Fair, and at the midway games, you come across the one rare, crooked carnival game. In this game, ba

lls have to be launched into holes in the wall. Placing the ball in the topmost hole wins you a giant cuddly bear, so you decide to aim for it. The wall is placed 2 m away from the launcher, and the top hole is 1.5 m above the launcher. The holes are designed in such a way that the ball will only go through if it is travelling horizontally when it reaches the hole. You can adjust the launch speed and angle of the ball. What settings for the speed and angle (measured from the horizontal) should you set the launcher in order to win the cuddly bear.
Physics
1 answer:
Anettt [7]4 years ago
4 0

Answer:

Explanation:

At the top , horizontal displacement = 2 m .

vertical displacement = 1.5 m .

Let time to reach the hole be t . Let the ball is launched at angle θ with horizontal with velocity u .

ucosθ x t = 2

from v = u - 2gt

0 = usinθ - 2gt

usinθ =  2gt

v² = u² - 2gh

u²sin²θ = 2 g h

u²sin²θ = 2 x 9.8 x 1.5

usinθ =  5.42

usinθ =  2gt

2gt = 5.42

t = .2765 s

ucosθ x t = 2

ucosθ x .2765 = 2

ucosθ = 7.23

usinθ =  5.42

dividing ,

Tanθ = .75

θ = 37°

usin37 =  5.42

u = 9.03 m /s

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Explanation :

It is given that,

Diameter of the coil, d = 20 cm = 0.2 m

Radius of the coil, r = 0.1 m

Number of turns, N = 3000

Induced EMF, \epsilon=1.5\ V

Magnitude of Earth's field, B=10^{-4}\ T

We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :

\epsilon=NBA\omega

\omega=\dfrac{\epsilon}{NBA}

\omega=\dfrac{1.5}{3000\times 10^{-4}\times \pi (0.1)^2}

\omega=159.15\ rad/s

So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.

3 0
3 years ago
A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse t
WARRIOR [948]

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

          θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

         θ = 1.22 550 10⁻⁹ / 0.002

         θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

       tan θ = y / L

       y = L tan θ

       y = 2,389 10⁵ tan 3,355 10⁻⁴

       y = 8.02 10¹ mi

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This is the smallest size of an object seen directly by the eye

5 0
4 years ago
stephen buys a new moped . he travels 4km south and then 6km east. how far does he need to go to get back where he started??
stiks02 [169]

Answer:

My answer is 7.2 km

Explanation:

When Stephen goes to the south and then to the east, he is drawing a right triangle, where the 4 km and 6 km sides are the cathetus of a right triangle.

Then we use the Pithagorean theorem to solve this problem. We need to find the hypotenuse.

c² = a² + b²

c² = 4² + 6²

c² = 16 + 36

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c = 7.2 km

7 0
3 years ago
It is rate for any motion to
Gnoma [55]

Answer:

a. stay the same for very long

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It is rare for any motion to stay the same for a very long time. The force applied on a body causes changes in the magnitude of motion.

  • For motion to remain constant, there must not be a net force acting on the body
  • All the forces on the body must be balanced.
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3 0
3 years ago
A pear hangs in a tree at a height of 1.8 m. The pear has a mass of 0.2 kg. The pear falls out of the tree and lands on the grou
TiliK225 [7]

a) PE=mgh=0.2*9.8*1.2=2.352 J

b) KE=PE=2.352 J

c) v=\sqrt{\frac{2KE}{m}}=4.85 m/s

6 0
4 years ago
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